E9 A.

asked • 08/11/15

Finding the amount of ways and its probability:

A shipment of 250 netbooks contains 3 defective units. Determine how many ways a vending company can buy three of these units and receive
(a) no defective units.
(b) all defective units.
(c) at least one good unit.
Now, find the probability of the vending company receiving
(a) no defective units.
(b) all defective units.
(c) at least one good unit.

1 Expert Answer

By:

Bill P. answered • 08/11/15

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I will address the last part here (The probabilities of receiving each of the options listed:
 
P(no defective) = the product of the 3 fractions 247/250   246/249    245/248  which is 14,886,690/15,438,000
 
As a decimal this is approximately 0.9643
 
P (all defective) is the product of the 3 fractions 3/250   2/249   1/248    which is   6/15,438,000 or 1/2,573,000
 
As a decimal this is 0.0000003887 or 0.0000004
 
P (at least one good) is equal to 1 minus the answer to p (all defective).  As a fraction this would be 2,572,999/2,573,000 which is also 0.9999996113
 
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Bill P.

I will now address the first part:
 
the number of ways that a company could receive:
 
a) no defective items when purchasing 3 items is (247*246*245)/ (3*2*1)  =  2,481,115  many ways
 
b) all defective items when purchasing 3 items is (3 * 2 * 1 )/ (3 * 2 * 1) = 1 (obviously) way (Please realize that there are only 3 defective units. So, of course there is only one way to get all three of them.
 
C) At least one good unit when purchasing 3 items is (250*249*248) / (3*2*1) - 1 = 2,572,999 ways
 
There are 2,573,000 unique ways to obtain 3 items from 250 items. In every case except one (when all three are defective) you have at least one that is not defective. Therefore 2,573,000 - 1 = 2,572,999 unique ways to have 3 items with at least one good one.
 
c)
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08/11/15

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